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########################################################################

## -*- texinfo -*-
## @deftypefn  {} {[@var{r}, @var{p}, @var{k}, @var{e}] =} residue (@var{b}, @var{a})
## @deftypefnx {} {[@var{b}, @var{a}] =} residue (@var{r}, @var{p}, @var{k})
## @deftypefnx {} {[@var{b}, @var{a}] =} residue (@var{r}, @var{p}, @var{k}, @var{e})
## The first calling form computes the partial fraction expansion for the
## quotient of the polynomials, @var{b} and @var{a}.
##
## The quotient is defined as
## @tex
## $$
## {B(s)\over A(s)} = \sum_{m=1}^M {r_m\over (s-p_m)^e_m}
##   + \sum_{i=1}^N k_i s^{N-i}.
## $$
## @end tex
## @ifnottex
##
## @example
## @group
## B(s)    M       r(m)        N
## ---- = SUM ------------- + SUM k(i)*s^(N-i)
## A(s)   m=1 (s-p(m))^e(m)   i=1
## @end group
## @end example
##
## @end ifnottex
## @noindent
## where @math{M} is the number of poles (the length of the @var{r}, @var{p},
## and @var{e}), the @var{k} vector is a polynomial of order @math{N-1}
## representing the direct contribution, and the @var{e} vector specifies the
## multiplicity of the m-th residue's pole.
##
## For example,
##
## @example
## @group
## b = [1, 1, 1];
## a = [1, -5, 8, -4];
## [r, p, k, e] = residue (b, a)
##    @result{} r = [-2; 7; 3]
##    @result{} p = [2; 2; 1]
##    @result{} k = [](0x0)
##    @result{} e = [1; 2; 1]
## @end group
## @end example
##
## @noindent
## which represents the following partial fraction expansion
## @tex
## $$
## {s^2+s+1\over s^3-5s^2+8s-4} = {-2\over s-2} + {7\over (s-2)^2} + {3\over s-1}
## $$
## @end tex
## @ifnottex
##
## @example
## @group
##         s^2 + s + 1       -2        7        3
##    ------------------- = ----- + ------- + -----
##    s^3 - 5s^2 + 8s - 4   (s-2)   (s-2)^2   (s-1)
## @end group
## @end example
##
## @end ifnottex
##
## The second calling form performs the inverse operation and computes the
## reconstituted quotient of polynomials, @var{b}(s)/@var{a}(s), from the
## partial fraction expansion; represented by the residues, poles, and a direct
## polynomial specified by @var{r}, @var{p} and @var{k}, and the pole
## multiplicity @var{e}.
##
## If the multiplicity, @var{e}, is not explicitly specified the multiplicity
## is determined by the function @code{mpoles}.
##
## For example:
##
## @example
## @group
## r = [-2; 7; 3];
## p = [2; 2; 1];
## k = [1, 0];
## [b, a] = residue (r, p, k)
##    @result{} b = [1, -5, 9, -3, 1]
##    @result{} a = [1, -5, 8, -4]
##
## where mpoles is used to determine e = [1; 2; 1]
## @end group
## @end example
##
## Alternatively the multiplicity may be defined explicitly, for example,
##
## @example
## @group
## r = [7; 3; -2];
## p = [2; 1; 2];
## k = [1, 0];
## e = [2; 1; 1];
## [b, a] = residue (r, p, k, e)
##    @result{} b = [1, -5, 9, -3, 1]
##    @result{} a = [1, -5, 8, -4]
## @end group
## @end example
##
## @noindent
## which represents the following partial fraction expansion
## @tex
## $$
## {-2\over s-2} + {7\over (s-2)^2} + {3\over s-1} + s = {s^4-5s^3+9s^2-3s+1\over s^3-5s^2+8s-4}
## $$
## @end tex
## @ifnottex
##
## @example
## @group
##  -2        7        3         s^4 - 5s^3 + 9s^2 - 3s + 1
## ----- + ------- + ----- + s = --------------------------
## (s-2)   (s-2)^2   (s-1)          s^3 - 5s^2 + 8s - 4
## @end group
## @end example
##
## @end ifnottex
## @seealso{mpoles, poly, roots, conv, deconv}
## @end deftypefn

function [r, p, k, e] = residue (b, a, varargin)

  if (nargin < 2 || nargin > 4)
    print_usage ();
  endif

  tol = .001;

  if (nargin >= 3)
    if (nargin >= 4)
      e = varargin{2};
    else
      e = [];
    endif
    ## The inputs are the residue, pole, and direct part.
    ## Solve for the corresponding numerator and denominator polynomials.
    [r, p] = rresidue (b, a, varargin{1}, tol, e);
    return;
  endif

  ## Make sure both polynomials are in reduced form, and scaled.
  a = polyreduce (a);
  b = polyreduce (b);

  b /= a(1);
  a /= a(1);

  la = length (a);
  lb = length (b);

  ## Handle special cases here.
  if (la == 0 || lb == 0)
    k = r = p = e = [];
    return;
  elseif (la == 1)
    k = b / a;
    r = p = e = [];
    return;
  endif

  ## Find the poles.
  p = roots (a);
  lp = length (p);

  ## Sort poles so that multiplicity loop will work.
  [e, idx] = mpoles (p, tol, 1);
  p = p(idx);

  ## For each group of pole multiplicity, set the value of each
  ## pole to the average of the group.  This reduces the error in
  ## the resulting poles.
  p_group = cumsum (e == 1);
  for ng = 1:p_group(end)
    m = find (p_group == ng);
    p(m) = mean (p(m));
  endfor

  ## Find the direct term if there is one.
  if (lb >= la)
    ## Also return the reduced numerator.
    [k, b] = deconv (b, a);
    lb = length (b);
  else
    k = [];
  endif

  ## Determine if the poles are (effectively) zero.
  small = max (abs (p));
  if (isa (a, "single") || isa (b, "single"))
    small = max ([small, 1]) * eps ("single") * 1e4 * (1 + numel (p))^2;
  else
    small = max ([small, 1]) * eps * 1e4 * (1 + numel (p))^2;
  endif
  p(abs (p) < small) = 0;

  ## Determine if the poles are (effectively) real, or imaginary.
  idx = (abs (imag (p)) < small);
  p(idx) = real (p(idx));
  idx = (abs (real (p)) < small);
  p(idx) = 1i * imag (p(idx));

  ## The remainder determines the residues.  The case of one pole is trivial.
  if (lp == 1)
    r = polyval (b, p);
    return;
  endif

  ## Determine the order of the denominator and remaining numerator.
  ## With the direct term removed, the potential order of the numerator
  ## is one less than the order of the denominator.
  aorder = numel (a) - 1;
  border = aorder - 1;

  ## Construct a system of equations relating the individual
  ## contributions from each residue to the complete numerator.
  A = zeros (border+1, border+1);
  B = prepad (reshape (b, [numel(b), 1]), border+1, 0);
  for ip = 1:numel (p)
    ri = zeros (size (p));
    ri(ip) = 1;
    A(:,ip) = prepad (rresidue (ri, p, [], tol), border+1, 0).';
  endfor

  ## Solve for the residues.
  ## FIXME: Use a pre-conditioner d to make A \ B work better (bug #53869).
  ##        It would be better to construct A and B so they are not close to
  ##        singular in the first place.
  d = max (abs (A), [], 2);
  r = (diag (d) \ A) \ (B ./ d);

endfunction

## Reconstitute the numerator and denominator polynomials
## from the residues, poles, and direct term.
function [pnum, pden, e] = rresidue (r, p, k = [], tol = [], e = [])

  if (! isempty (e))
    idx = 1:numel (p);
  else
    [e, idx] = mpoles (p, tol, 0);
    p = p(idx);
    r = r(idx);
  endif

  idx = 1:numel (p);
  for n = idx
    pn = [1, -p(n)];
    if (n == 1)
      pden = pn;
    else
      pden = conv (pden, pn);
    endif
  endfor

  ## D is the order of the denominator
  ## K is the order of the direct polynomial
  ## N is the order of the resulting numerator
  ## pnum(1:(N+1)) is the numerator's polynomial
  ## pden(1:(D+1)) is the denominator's polynomial
  ## pm is the multiple pole for the nth residue
  ## pn is the numerator contribution for the nth residue

  D = numel (pden) - 1;
  K = numel (k) - 1;
  N = K + D;
  pnum = zeros (1, N+1);
  for n = idx(abs (r) > 0)
    p1 = [1, -p(n)];
    pn = 1;
    for j = 1:n - 1
      pn = conv (pn, [1, -p(j)]);
    endfor
    for j = n + 1:numel (p)
      pn = conv (pn, [1, -p(j)]);
    endfor
    for j = 1:e(n) - 1
      pn = deconv (pn, p1);
    endfor
    pn = r(n) * pn;
    pnum += prepad (pn, N+1, 0, 2);
  endfor

  ## Add the direct term.
  if (numel (k))
    pnum += conv (pden, k);
  endif

  pnum = polyreduce (pnum);
  pden = polyreduce (pden);

endfunction


%!test
%! b = [1, 1, 1];
%! a = [1, -5, 8, -4];
%! [r, p, k, e] = residue (b, a);
%! assert (r, [-2; 7; 3], 1e-12);
%! assert (p, [2; 2; 1], 1e-12);
%! assert (isempty (k));
%! assert (e, [1; 2; 1]);
%! k = [1 0];
%! b = conv (k, a) + prepad (b, numel (k) + numel (a) - 1, 0);
%! a = a;
%! [br, ar] = residue (r, p, k);
%! assert (br, b, 1e-12);
%! assert (ar, a, 1e-12);
%! [br, ar] = residue (r, p, k, e);
%! assert (br, b, 1e-12);
%! assert (ar, a, 1e-12);

%!test
%! b = [1, 0, 1];
%! a = [1, 0, 18, 0, 81];
%! [r, p, k, e] = residue (b, a);
%! r1 = [-5i; 12; +5i; 12]/54;
%! p1 = [+3i; +3i; -3i; -3i];
%! assert (r, r1, 1e-12);
%! assert (p, p1, 1e-12);
%! assert (isempty (k));
%! assert (e, [1; 2; 1; 2]);
%! [br, ar] = residue (r, p, k);
%! assert (br, b, 1e-12);
%! assert (ar, a, 1e-12);

%!test
%! r = [7; 3; -2];
%! p = [2; 1; 2];
%! k = [1 0];
%! e = [2; 1; 1];
%! [b, a] = residue (r, p, k, e);
%! assert (b, [1, -5, 9, -3, 1], 1e-12);
%! assert (a, [1, -5, 8, -4], 1e-12);
%! [rr, pr, kr, er] = residue (b, a);
%! [~, m] = mpoles (rr);
%! [~, n] = mpoles (r);
%! assert (rr(m), r(n), 1e-12);
%! assert (pr(m), p(n), 1e-12);
%! assert (kr, k, 1e-12);
%! assert (er(m), e(n), 1e-12);

%!test
%! b = [1];
%! a = [1, 10, 25];
%! [r, p, k, e] = residue (b, a);
%! r1 = [0; 1];
%! p1 = [-5; -5];
%! assert (r, r1, 1e-12);
%! assert (p, p1, 1e-12);
%! assert (isempty (k));
%! assert (e, [1; 2]);
%! [br, ar] = residue (r, p, k);
%! assert (br, b, 1e-12);
%! assert (ar, a, 1e-12);

## The following test is due to Bernard Grung
%!test <*34266>
%! z1 =  7.0372976777e6;
%! p1 = -3.1415926536e9;
%! p2 = -4.9964813512e8;
%! r1 = -(1 + z1/p1)/(1 - p1/p2)/p2/p1;
%! r2 = -(1 + z1/p2)/(1 - p2/p1)/p2/p1;
%! r3 = (1 + (p2 + p1)/p2/p1*z1)/p2/p1;
%! r4 = z1/p2/p1;
%! r = [r1; r2; r3; r4];
%! p = [p1; p2; 0; 0];
%! k = [];
%! e = [1; 1; 1; 2];
%! b = [1, z1];
%! a = [1, -(p1 + p2), p1*p2, 0, 0];
%! [br, ar] = residue (r, p, k, e);
%! assert (br, [0,0,b], 1e-7);
%! assert (ar, a, 1e-8);

%!test <*49291>
%! rf = [1e3, 2e3, 1e3, 2e3];
%! cf = [316.2e-9, 50e-9, 31.6e-9, 5e-9];
%! [num, den] = residue (1./cf,-1./(rf.*cf),0);
%! assert (numel (num), 4);
%! assert (numel (den), 5);
%! assert (den(1), 1);

%!test <*51148>
%! r = [1.0000e+18, 3.5714e+12, 2.2222e+11, 2.1739e+10];
%! pin = [-1.9231e+15, -1.6234e+09, -4.1152e+07, -1.8116e+06];
%! k = 0;
%! [p, q] = residue (r, pin, k);
%! assert (p(4), 4.6828e+42, -1e-5);

%!test <*60384>
%! B = [1315.789473684211];
%! A = [1, 1.100000536842105e+04, 1.703789473684211e+03, 0];
%! poles1 = roots (A);
%! [r, p, k, e] = residue (B, A);
%! [B1, A1] = residue (r, p, k, e);
%! assert (B, B1);
%! assert (A, A1);
